solved errors

Author: ajay-dhangar

dsa-solutions/lc-solutions/0000-0099/0004-Median-of-two-Sorted-Array.md

@@ -16,7 +16,7 @@ In this page, we will solve the problem [Median of Two Sorted Arrays](https://le
 
 Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return the median of the two sorted arrays.
 
-The overall run time complexity should be <code>$O(log (m+n))$</code>.
+The overall run time complexity should be $O(log (m+n))$.
 
 ### Example 1
 
@@ -429,7 +429,7 @@ Let's test the solution with the sample test cases:
 Input: nums1 = [1, 3], nums2 = [2]
 Output: 2.00000
 Explanation: merged array = [1, 2, 3] and median is 2.
-````
+```
 </TabItem>
 
 <TabItem value="TestCase2" label="Case 2">
@@ -770,8 +770,7 @@ Let's test the solution with the sample test cases:
 Input: nums1 = [1, 3], nums2 = [2]
 Output: 2.00000
 Explanation: merged array = [1, 2, 3] and median is 2.
-````
-
+```
 </TabItem>
 <TabItem value="TestCase2" label="Case 2">
 ```plaintext
@@ -786,7 +785,7 @@ Explanation: merged array = [1, 2, 3, 4] and median is (2 + 3) / 2 = 2.5.
 
 :::info
 
-**Note**: The binary search approach is more efficient than the divide and conquer approach as it has a better time complexity of <code>$O(log(min(n, m)))$</code> compared to the divide and conquer approach. However, both approaches provide a solution to the problem of finding the median of two sorted arrays.
+**Note**: The binary search approach is more efficient than the divide and conquer approach as it has a better time complexity of $O(log(min(n, m)))$ compared to the divide and conquer approach. However, both approaches provide a solution to the problem of finding the median of two sorted arrays.
 
 **Which approach is best for you?**
 
@@ -795,7 +794,7 @@ The binary search approach is more efficient and recommended for solving the pro
 :::
 
 :::tip
-When asked to find the median of two sorted arrays, a direct approach that merges the two arrays and then finds the median will work but isn't optimal. Given the problem's constraints, we can leverage the fact that the arrays are already sorted and use binary search to find the median in $$ O(\log(\min(n, m))) $$ time complexity.
+When asked to find the median of two sorted arrays, a direct approach that merges the two arrays and then finds the median will work but isn't optimal. Given the problem's constraints, we can leverage the fact that the arrays are already sorted and use binary search to find the median in $O(\log(\min(n, m)))$ time complexity.
 
 The key idea is to use binary search to partition the smaller array in such a way that we can easily find the median by comparing elements around the partition.
 
@@ -823,19 +822,19 @@ The key idea is to use binary search to partition the smaller array in such a wa
 
 4. **Boundary Conditions**:
 
-   - Handle cases where partitions might go out of bounds. If $$ \text{partitionX} $$ is 0, it means there are no elements on the left side of $$ \text{nums1} $$. If $$ \text{partitionX} $$ is $$ x $$, it means there are no elements on the right side of $$ \text{nums1} $$.
+   - Handle cases where partitions might go out of bounds. If $$ \text{partitionX} $$ is 0, it means there are no elements on the left side of $$ \text{nums1} $$. If $$ \text{partitionX} $$ is $x$, it means there are no elements on the right side of $$ \text{nums1} $$.
 
 5. **Check Valid Partition**:
 
    - A valid partition is one where the maximum element on the left side of both partitions is less than or equal to the minimum element on the right side of both partitions:
      $$
      \text{maxX} \leq \text{minY} \quad \text{and} \quad \text{maxY} \leq \text{minX}
      $$
-     Here, $$ \text{maxX} $$ is the largest element on the left side of $$ \text{nums1} $$, $$ \text{minX} $$ is the smallest element on the right side of $$ \text{nums1} $$, and similarly for $$ \text{nums2} $$.
+     Here, $\text{maxX}$ is the largest element on the left side of $\text{nums1}$, $\text{minX}$ is the smallest element on the right side of $\text{nums1}$, and similarly for $\text{nums2}$.
 
 6. **Calculate the Median**:
 
-   - If the total number of elements $$ (x + y) $$ is even, the median is the average of the two middle values:
+   - If the total number of elements $(x + y)$ is even, the median is the average of the two middle values:
      $$
      \text{median} = \frac{\text{max(maxX, maxY)} + \text{min(minX, minY)}}{2}
      $$
@@ -845,8 +844,8 @@ The key idea is to use binary search to partition the smaller array in such a wa
      $$
 
 7. **Adjust Binary Search**:
-   - If $$ \text{maxX} > \text{minY} $$, it means we need to move the partition in $$ \text{nums1} $$ to the left, so adjust $$ \text{high} $$.
-   - If $$ \text{maxY} > \text{minX} $$, it means we need to move the partition in $$ \text{nums1} $$ to the right, so adjust $$ \text{low} $$.
+   - If $\text{maxX} > \text{minY}$, it means we need to move the partition in $\text{nums1}$ to the left, so adjust $\text{high}$.
+   - If $\text{maxY} > \text{minX}$, it means we need to move the partition in $\text{nums1}$ to the right, so adjust $\text{low}$.
 
 :::